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Q9(ii):

In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that: (ii) $\frac{CA}{PA} = \frac{BC}{MP}$

Solution :

Given: Two right-angled triangles $\triangle ABC$ and $\triangle AMP$.

1. $\angle ABC = 90^\circ$ (Given that $\triangle ABC$ is right-angled at $B$).

2. $\angle AMP = 90^\circ$ (Given that $\triangle AMP$ is right-angled at $M$).

To Prove: $\frac{CA}{PA} = \frac{BC}{MP}$

B C A M P

Step 1: Identifying the triangles to be compared.

To prove the ratio $\frac{CA}{PA} = \frac{BC}{MP}$, we must establish the similarity between $\triangle ABC$ and $\triangle AMP$.

Step 2: Analyzing the angles of $\triangle ABC$ and $\triangle AMP$.

In $\triangle ABC$ and $\triangle AMP$:

1. $\angle ABC = \angle AMP = 90^\circ$ (Given, both are right angles).

2. $\angle BAC = \angle MAP$ (This is a common angle shared by both triangles, as vertex $A$ is common to both).

Step 3: Applying the AA (Angle-Angle) Similarity Criterion.

The AA Similarity Criterion states that if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Since $\angle ABC = \angle AMP$ and $\angle BAC = \angle MAP$, we conclude:

$\triangle ABC \sim \triangle AMP$ [By AA Similarity Criterion]

Step 4: Using the property of similar triangles.

When two triangles are similar, their corresponding sides are proportional.

For $\triangle ABC \sim \triangle AMP$, the ratios of the corresponding sides are:

$\frac{AB}{AM} = \frac{BC}{MP} = \frac{CA}{PA}$

Step 5: Extracting the required ratio.

From the proportionality established in Step 4, we specifically select the equality:

$\frac{CA}{PA} = \frac{BC}{MP}$

Final Answer: Since $\triangle ABC \sim \triangle AMP$ by the AA similarity criterion, the ratio of their corresponding sides is equal, hence $\frac{CA}{PA} = \frac{BC}{MP}$ is proved.


More Questions from Class 10 Mathematics Triangles EXERCISE 6.3


CBSE Solutions for Class 10 Mathematics Triangles


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